Question : If $a=\frac{1}{a-\sqrt{6}}$ and $(a>0)$, then the value of $\left(a+\frac{1}{a}\right)$ is:
Option 1: $\sqrt{6}$
Option 2: $\sqrt{10}$
Option 3: $\sqrt{15}$
Option 4: $\sqrt{7}$
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Correct Answer: $\sqrt{10}$
Solution : $a=\frac{1}{a-\sqrt{6}}(x>0)$ $⇒a^2-a\sqrt 6-1=0$ $⇒a^2-1=a\sqrt 6$ Multiplying both sides by $\frac{1}{a}$, we get, $⇒a-\frac{1}{a}=\sqrt 6$ Squaring both sides, we get, $⇒(a-\frac{1}{a})^2=(\sqrt 6)^2$ $⇒a^2+\frac{1}{a^2}-2=6$ Adding 4 to both sides, we get, $⇒a^2+\frac{1}{a^2}+2=6+4$ $⇒(a+\frac{1}{a})^2=10$ $\therefore a+\frac{1}{a}=\sqrt{10}$ Hence, the correct answer is $\sqrt{10}$.
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